Question: A scientist measures the initial amount of Carbon- $14$ in a substance to be $25$ grams. The relationship between $A$, the amount of Carbon- $14$ remaining in that substance, in grams, and $t$, the elapsed time, in years, since the initial measurement is modeled by the following equation. $A=25e^{-0.00012t}$ In how many years will the substance contain exactly $20$ grams $(\text{g})$ of Carbon- $14$ ? Give an exact answer expressed as a natural logarithm. years
Explanation: Thinking about the problem We want to know how many years, $t$, it will take for the amount of Carbon- $14$ remaining, $A$, to be $20\text{ g}$. So we need to find the value of $t$ for which $A=20$. Substituting $20$ in for $A$ in the model gives us the following equation: $20=25\cdot e^{-0.00012t}$ Solving the equation We can solve the equation as shown below. $\begin{aligned}25\cdot e^{-0.00012t}&=20\\\\ e^{-0.00012t}&=0.8\\\\ -0.00012t&=\ln\left(0.8\right)\\\\ t&=\dfrac{{\,\ln\left(0.8\right)}}{-0.00012}\\\\ \end{aligned}$ It will take $\dfrac{{\,\ln\left(0.8\right)}}{-0.00012}$ years for the amount of Carbon- $14$ remaining in the substance to be $20\text{ g}$. The expression above represents an exact solution to the problem. We can use a calculator to approximate the value of the expression, but this will be a rounded inexact answer. The answer The answer is $\dfrac{{\,\ln\left(0.8\right)}}{-0.00012}$ years.